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Algebra - Surds & Indices - Previous Year CAT/MBA Questions

The best way to prepare for Algebra - Surds & Indices is by going through the previous year Algebra - Surds & Indices omet questions. Here we bring you all previous year Algebra - Surds & Indices omet questions along with detailed solutions.

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It would be best if you clear your concepts before you practice previous year Algebra - Surds & Indices omet questions.

1. CMAT 2024 Slot 2 QA | Algebra - Surds & Indices omet Question

Arrange the following surds in increasing order :

(A) $\sqrt[3]{2}$
(B) $\sqrt{3}$
(C) $\sqrt[4]{5}$
(D) $\sqrt[6]{7}$

Choose the correct answer from the options given below :

(a)
A < B < C < D
(b)
D < A < C < B
(c)
D < B < C < A
(d)
A < D < C < B

2. CMAT 2024 Slot 1 QA | Algebra - Surds & Indices omet Question

Choose the correct answer from the options given below:

(a)
(A) - (I), (B) - (IV), (C) - (III), (D) - (II)
(b)
(A) - (I), (B) - (III), (C) - (IV), (D - (II)
(c)
(A) - (II), (B) - (IV), (C) - (I), (D) - (III)
(d)
(A) - (II), (B) - (I), (C) - (IV), (D) - (III)

3. CMAT 2024 Slot 1 QA | Algebra - Surds & Indices omet Question

Arrange the following in increasign order:

A) $\sqrt[4]{3}$
B) $\sqrt[3]{2}$
C) $\sqrt[6]{5}$
D) $\sqrt[6]{7}$

Choose the correct answer from the options given below:

(a)
A < D < B < C
(b)
B < C < A < D
(c)
A < C < B < D
(d)
B < A < C < D

4. XAT 2019 QADI | Algebra - Surds & Indices omet Question

If $\sqrt[3]{7^{a} \times (35)^{(b + 1)} \times (20)^{(c + 2)}}$ is a whole number then which one of the statements below is consistent with it?

(a)
a = 3, b = 2, c = 1
(b)
a = 3, b = 1, c = 1
(c)
a = 2, b = 1, c = 2
(d)
a = 2, b = 1, c = 1
(e)
a = 1, b = 2, c = 2

5. IIFT 2019 QA | Algebra - Surds & Indices omet Question

If x = 8 - $\sqrt{32}$ and y = 2 + √2, then ${(x + \frac{1}{y})}^{2}$ is given by:

(a)
16x²/25
(b)
64x²/81
(c)
25y²/16
(d)
81x²/64

Answer: Option D

Text Explanation :

x = 8 − √32 = 8 − 4√2 = 4(2 − √2)

$\frac{1}{y}$ = $\frac{1}{2 + \sqrt{2}}$

On rationalising, we get

$\frac{1}{y}$ = $\frac{1}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}}$ = $\frac{2 - \sqrt{2}}{2}$ = $\frac{x}{8}$

∴ x + 1/y = x + x/8 = 9x/8

⇒ ${(x + \frac{1}{y})}^{2}$ = ${(x + \frac{x}{8})}^{2}$ = ${(\frac{9 x}{8})}^{2}$ = $\frac{81 x^{2}}{64}$

Hence option (d).

6. IIFT 2016 QA | Algebra - Surds & Indices omet Question

The highest number amongst $\sqrt{2}, \sqrt[3]{3},$ and $\sqrt[4]{4}$

(a)
$\sqrt{2}$
(b)
$\sqrt[3]{3}$
(c)
$\sqrt[4]{4}$
(d)
None of the above

7. IIFT 2015 QA | Algebra - Surds & Indices omet Question

The value of x for which the equation $\sqrt{4 x - 9} + \sqrt{4 x + 9} = 5 + \sqrt{7}$ will be satisfied is:

(a)
1
(b)
2
(c)
3
(d)
4

8. IIFT 2015 QA | Algebra - Surds & Indices omet Question

The simplest value of the expression ${\{\frac{4^{p + \frac{1}{4}} \times \sqrt{2 \times 2^{p}}}{2 \times \sqrt{2^{- p}}}\}}^{\frac{1}{p}}$ is:

(a)
4
(b)
8
(c)
4^p
(d)
8^p

Answer: Option B

Text Explanation :

$\{\frac{4^{p + \frac{1}{4}} \times \sqrt{2 \times 2^{p}}}{2 \times \sqrt{2^{- p}}}\} = \{\frac{4^{p} \times 4^{\frac{1}{4}} \times \sqrt{2} \times \sqrt{2^{p}}}{2 \times \sqrt{2^{- p}}}\} = \{\frac{2^{2 p} \times \sqrt{2} \times \sqrt{2} \times \sqrt{2^{p}}}{2 \times \sqrt{2^{- p}}}\} = 2^{2 p} \times \sqrt{2^{p}} \times \sqrt{2^{p}} = 2^{3 p}$

${\{\frac{4^{p + \frac{1}{4}} \times \sqrt{2 \times 2^{p}}}{2 \times \sqrt{2^{- p}}}\}}^{\frac{1}{p}} = {(2^{3 p})}^{\frac{1}{p}} = 8$

Hence, option (b).